# The Elements of Euclid for the Use of Schools and Colleges/Book I.

EUCLID'S ELEMENTS.

*BOOK I.*

DEFINITIONS.

1. A point is that which has no parts, or which has no magnitude.

2. A line is length without breadth.

3. The extremities of a line are points.

4. A straight line is that which lies evenly between its extreme points.

5. A superficies is that which has only length and breadth.

6. The extremities of a superficies are lines.

7. A plane superficies is that in which any two points being taken, the straight line between them lies wholly in that superficies.

8. A plane angle is the inclination of two lines to one another in a plane, which meet together, but are not in the same direction.

9. A plane rectilineal angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line.

*Note*. When several angles are at one point

*B*, any one of them is expressed by three letters, of which the letter which is at the vertex of the angle, that is, at the point at which the straight lines that contain the angle meet one another, is put between the other two letters, and one of these two letters is somewhere on one of those straight lines, and the other letter on the other straight line. Thus, the angle which is contained by the straight lines

*AB*,

*CB*is named the angle

*ABC*, or

*CBA*; the angle which is contained by the straight lines

*AB*,

*DB*is named the angle

*ABD*, or

*DBA*; and the angle which is contained by the straight lines

*DB*,

*CB*is named the angle

*DBC*, or

*CBD*; but if there be only one angle at a point, it may be expressed by a letter placed at that point; as the angle at

*E*.

10. When a straight line standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it.

11. An obtuse angle is that which is greater than a right angle.

12. An acute angle is that which is less than a right angle.

13. A term or boundary is the extremity of any thing.

14. A figure is that which is enclosed by one or more boundaries.15. A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another:

16. And this point is called the centre of the circle.

17. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.

[A radius of a circle is a straight line drawn from the centre to the circumference.]

18. A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter.

19. A segment of a circle is the figure contained by a straight line and the circumference which it cuts off.

20. Rectilineal figures are those which are contained by straight lines:

21. Trilateral figures, or triangles, by three straight lines:

22. Quadrilateral figures by four straight lines:

23. Multilateral figures, or polygons, by more than four straight lines.24. Of three-sided figures,

An equilateral triangle is that which has three equal sides:

25. An isosceles triangle is that which has two sides equal:26. A scalene triangle is that which has three unequal sides:

27. A right-angled triangle is that which has a right angle:

[The side opposite to the right angle in a right-angled triangle is frequently called the hypotenuse.]

28. An obtuse-angled triangle is that which has an obtuse angle:

29. An acute-angled triangle is that which has three acute angles.

Of four-sided figures,

30. A square is that which has all its sides equal, and all its angles right angles:

31. An oblong is that which has all its angles right angles, but not all its sides equal:

32. A rhombus is that which has all its sides equal, but its angles are not right angles:

33. A rhomboid is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles:34. All other four-sided figures besides these are called trapeziums.

35. Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways do not meet.[*Note*. The terms *oblong* and *rhomboid* are not often used. Practically the following definitions are used. Any four-sided figure is called a *quadrilateral*. A line joining two opposite angles of a quadrilateral is called a *diagonal*. A quadrilateral which has its opposite sides parallel is called a *parallelogram'*. The words *square* and *rhombus* are used in the sense defined by Euclid; and the word *rectangle* is used instead of the word *oblong*.

Some writers propose to restrict the word *trapezium* to a quadrilateral which has two of its sides parallel; and it would certainly be convenient if this restriction were universally adopted.]

POSTULATES.

Let it be granted,

1. That a straight line may be drawn from any one point to any other point:

2. That a terminated straight line may be produced to any length in a straight line:

3. And that a circle may be described from any centre, at any distance from that centre.

AXIOMS.

1. Things which are equal to the same thing are equal to one another.

2. If equals be added to equals the wholes are equal.

3. If equals be taken from equals the remainders are equal.

4. If equals be added to unequals the wholes are unequal.

5. If equals be taken from unequals the remainders are unequal.

6. Things which are double of the same thing are equal to one another.

7. Things which are halves of the same thing are equal to one another.

8. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.

9. The whole is greater than its part.

10. Two straight lines cannot enclose a space.

11. All right angles are equal to one another.

12. If a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these straight lines, being continually produced, shall at length meet on that side on which are the angles which are less than two right angles.

PROPOSITION 1. *PROBLEM.*

To describe an equilateral triangle on a given finite straight line.

Let *AB* be the given straight line; it is required to describe an equilateral triangle on *AB*.

From the centre *A* at the distance *AB* describe the circle *BCD*. [*Postulate* 3.

From the centre *B*, at the distance *BA*, describe the circle *ACE*. [*Postulate* 3.

From the point *C*, at which the circles cut one another, draw the straight lines *CA* and *CB* to the points *A* and *B*. [*Post*. 1.

*ABC* shall be an equilateral triangle.

Because the point *A* is the centre of the circle *BCD*, *AC* is equal to *AB*. [*Definition* 15.

And because the point *B* is the centre of the circle *ACE*, *BC* is equal to *BA*. [*Definition* 15.

But it has been shewn that *CA* is equal to *AB*;

therefore *CA* and *CB* are each of them equal to *AB*.

But things which are equal to the same thing are equal to one another. [*Axiom* 1.

Therefore *CA* is equal to *CB*.

Therefore *CA*, *AB*, *BC* are equal to one another.

Wherefore the *triangle ABC is equilateral*, [*Def*. 24. *and it is described on the given straight line AB*. Q.E.F.

PROPOSITION 2. *PROBLEM.*

From a given point to draw a straight line equal to a given straight line.

Let *A* be the given point, and *BC* the given straight line: it is required to draw from the point *A* a straight line equal to *BC*.

From the point *A* to *B* draw the straight line *AB*; [*Post*. 1.

and on it describe the equilateral triangle *DAB*, [I. 1.

and produce the straight lines *DA*, *DB* to *E* and *F*. [*Post*. 2.

From the centre *B*, at the distance *BC*, describe the circle *CGH*, meeting *DF* at *G*. [*Post*. 3.

From the centre *D*, at the distance *DG*, describe the circle *GKL*, meeting *DE* at *L.* [*Post*. 3.*AL* shall be equal to *BC*.

Because the point *B* is the centre of the circle *CGH*, *BC* is equal to *BG*. [*Definition* 15.

And because the point *D* is the centre of the circle *GKL*, *DL* is equal to *DG*; [*Definition* 15.

and *DA*, *DB* parts of them are equal; [*Definition* 24.

therefore the remainder *AL* is equal to the remainder BG. [*Axiom* 3.

But it has been shewn that *BC* is equal to *BG*;

therefore *AL* and *BC* are each of them equal to *BG*.

But things which are equal to the same thing are equal to one another. [*Axiom* 1.

Therefore *AL* is equal to *BC*.

Wherefore *from the given point A a straight line AL has been drawn equal to the given straight line BC*. q.e.f.

PROPOSITION 3. *PROBLEM*.

From the greater of two given straight lines to cut off a part equal to the less.

Let *AB* and *C* be the two given straight lines, of which *AB* is the greater: it is required to cut off from *AB* the greater, a part equal to C the less.

From the point *A* draw the straight line *AD* equal to *C*; [I. 2.

and from the centre *A*, at the distance *AD*, describe the circle *DEF* meeting *AB* at *E*. [*Postulate* 3.*AE* shall be equal to *C*.

Because the point *A* is the centre of the circle *DEF*, *AE* is equal to *AD*. [*Definition* 15.

But *C* is equal to *AD*. [*Construction*.

Therefore *AE* and *C* are each of them equal to *AD*.

Therefore *AE* is equal to *C*. [*Axiom* 1.

Wherefore *from AB the greater of two given straight lines a part AE has been cut off equal to C the less.* q.e.f.

PROPOSITION 4. *THEOREM*.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases or third sides equal; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely those to which the equal sides are opposite.

Let *ABC*, *DEF* be two triangles which have the two sides *AB*, *AC* equal to the two sides *DE*, *DF*, each to each, namely, *AB* to *DE*, and *AC* to *DF*, and the angle *BAC* equal to the angle *EDF*: the base *BC* shall be equal to the base *EF*, and the triangle *ABC* to the triangle *DEF*, and the other angles shall be equal, each to each, to which the equal sides are opposite, namely, the angle *ABC* to the angle *DEF*, and the angle *ACB* to the angle *DFE*.

For if the triangle *ABC* be applied to the triangle *DEF*, so that the point *A* may be on the point *D*, and the straight line *AB* on the straight line *DE*, the point *B* will coincide with the point *E*, because *AB* is equal to *DE*, [*Hyp*.

And, *AB* coinciding with *DE*, *AC* will fall on *DF*, because the angle *BAC* is equal to the angle *EDF*. [*Hypothesis*.

Therefore also the point *C* will coincide with the point *F*, because *AC* is equal to *DF*. [*Hypothesis*.

But the point *B* was shewn to coincide with the point *E*, therefore the base *BC* will coincide with the base *EF*;

because, *B* coinciding with *E* and *C* with *F*, if the base *BC* does not coincide with the base *EF*, two straight lines will enclose a space; which is impossible. [*Axiom* 10.

Therefore the base *BC* coincides with the base *EF*, and is equal to it. [*Axiom* 8.

Therefore the whole triangle *ABC* coincides with the whole triangle *DEF*, and is equal to it. [*Axiom* 8.

And the other angles of the one coincide with the other angles of the other, and are equal to them, namely, the angle *ABC* to the angle *DEF*, and the angle *ACB* to the angle *DFE*.

Wherefore, *if two triangles* &c. q.e.d.

PROPOSITION 5. *THEOREM*.

The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another.

Let *ABC* be an isosceles triangle, having the side *AB* equal to the side *AC*, and let the straight lines *AB*, *AC* be produced to *D* and *E*: the angle *ABC* shall be equal to the angle *ACB*, and the angle *CBD* to the angle *BCE*.

In *BD* take any point *F*,

and from *AE* the greater cut off *AG* equal to AF the less, [I.3.

*FC*,

*GB*.

Because *AF* is equal to *AG* [*Constr.*

and *AB* to *AC*, [*Hypothesis*.

the two sides *FA*, *AC* are equal to the two sides *GA*, *AB*, each to each; and they contain the angle *FAG* common to the two triangles *AFC*, *AGB*;

therefore the base *FC* is equal to the base *GB*, and the triangle *AFC* to the triangle *AGB*, and the remaining angles of the one to the remaining angles of the other, each to each, to which the equal sides are opposite, namely the angle *ACF* to the angle *ABG*, and the angle *AFC* to the angle *AGB*. [I. 4.

And because the whole *AF* is equal to the whole *AG*, of which the parts *AB*, *AC* are equal, [*Hypothesis*.

the remainder *BF* is equal to the remainder *CG*. [*Axiom* 3.

And *FC* was shewn to be equal to *GB*;

therefore the two sides *BF*, *FC* are equal to the two sides *CG*, *GB*, each to each;

and the angle *BFC* was shewn to be equal to the angle *CGB*; therefore the triangles *BFC*, *CGB* are equal, and their other angles are equal, each to each, to which the equal sides are opposite, namely the angle *FBC* to the angle *GCB*, and the angle *BCF* to the angle *CBG*. [I. 4.

And since it has been shewn that the whole angle *ABG* is equal to the whole angle *ACF*,

and that the parts of these, the angles *CBG*, *BCF* are also equal;

therefore the remaining angle *ABC* is equal to the remaining angle *ACB*, which are the angles at the base of the triangle *ABC*. [*Axiom* 3.

And it has also been shewn that the angle *FBC* is equal to the angle *GCB*, which are the angles on the other side of the base.

Wherefore, *the angles* &c. q.e.d.

Corollary. Hence every equilateral triangle is also equiangular.

PROPOSITION 6. *THEOREM*.

If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another.

Let *ABC* be a triangle, having the angle *ABC* equal to the angle *ACB*: the side *AC* shall be equal to the side *AB*.

For if *AC* be not equal to *AB*, one of them must be greater than the other.

Let *AB* be the greater, and from it cut off *DB* equal to *AC* the less, [I. 3.

and join *DC*.

Then, because in the triangles *DBC*, *ACB*, *DB* is equal to *AC*, [*Construction*.

and *BC* is common to both,

the two sides *DB*, *BC* are equal to the two sides *AC*, *CB* each to each;

and the angle *DBC* is equal to the angle *ACB*; [*Hypothesis*.

therefore the base *DC* is equal to the base *AB*, and the triangle *DBC* is equal to the triangle *ACB*, [I. 4.

the less to the greater; which is absurd. [*Axiom* 9.

Therefore *AB* is not unequal to *AC*, that is, it is equal to it.

Wherefore, *if two angles* &c. q.e.d.

Corollary. Hence every equiangular triangle is also equilateral.

PROPOSITION 7. *THEOREM.*

On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity equal to one another.

If it be possible, on the same base *AB*, and on the same side of it, let there be two triangles *ACB*, *ADB*, having their sides *CA*, *DA*, which are terminated at the extremity *A* of the base, equal

*CB*,

*DB*, which are terminated at

*B*equal to one another.

Join *CD*. In the case in which the vertex of each triangle is without the other triangle;

because *AC* is equal to *AD*,[*Hypothesis*.

the angle *ACD* is equal to the angle *ADC*.[I. 5.

But the angle *ACD* is greater than the angle *BCD*,[*Ax*. 9.

therefore the angle *ADC* is also greater than the angle *BCD*;

much more then is the angle *BDC* greater than the angle *BCD*.

Again, because *BC* is equal to *BD*,[*Hypothesis*.

the angle *BDC* is equal to the angle *BCD*.[I. 5.

But it has been shewn to be greater; which is impossible.

But if one of the vertices as *D*, be within the other triangle *ACB*, produce *AC*, *AD* to *E*, *F*.

Then because *AC* is equal to *AD*, in the triangle *ACD*, [*Hyp*.

the angles *ECD*, *FDC*, on the other side of the base *CD*, are equal to one another. [I. 5.

But the angle *ECD* is greater than the angle *BCD*, [*Axiom* 9.

therefore the angle *FDC* is also greater than the angle *BCD*;

much more then is the angle *BDC* greater than the angle *BCD*.

Again, because *BC* is equal to *BD*, [*Hypothesis*.

the angle *BDC* is equal to the angle *BCD*. [I. 5.

But it has been shewn to be greater; which is impossible.

The case in which the vertex of one triangle is on a side of the other needs no demonstration.

Wherefore, *on the same base* &c. q.e.d.

PROPOSITION 8. *THEOREM*.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their *bases equal, the angle which is contained by the two sides of the one shall be equal to the angle which is contained by the two sides, equal to them, of the other.*

Let *ABC*, *DEF* be two triangles, having the two sides *AB*, *AC* equal to the two sides *DE*, *DF*, each to each, namely *AB* to *DE*, and *AC* to *DF*, and also the base *BC* equal to the base *EF*: the angle *BAC* shall be equal to the angle *EDF*.

For if the triangle *ABC* be applied to the triangle *DEF*, so that the point *B* may be on the point *E*, and the straight line *BC* on the straight line *EF*, the point *C* will also coincide with the point *F*, because *BC* is equal to *EF*. [*Hyp*.

Therefore, *BC* coinciding with *EF*, *BA* and *AC* will coincide with *ED* and *DF*.

For if the base *BC* coincides with the base *EF*, but the sides *BA*, *CA* do not coincide with the sides *ED*, *FD*, but have a different situation as *EG*, *FG*; then on the same base and on the same side of it there will be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise their sides which are terminated at the other extremity.

But this is impossible. [I. 7.

Therefore since the base *BC* coincides with the base *EF*, the sides *BA*, *AC* must coincide with the sides *ED*, *DF*. Therefore also the angle *BAC* coincides with the angle *EDF*, and is equal to it. [*Axiom* 8.

Wherefore, *if two triangles* &c. q.e.d.

PROPOSITION 9. *PROBLEM*.

To bisect a given rectilineal angle, that is to divide it into two equal angles.

Let*BAC*be the given rectilineal angle: it is required to bisect it.

Take any point *D* in *AB*, and from *AC* cut off *AE* equal to *AD*; [I. 3.

join *DE*, and on *DE*, on the side remote from *A*, describe the equilateral triangle *DEF*. [I. 1.

Join *AF*. The straight line *AF* shall bisect the angle *BAC*.

Because *AD* is equal to *AE*, [*Construction*.

and *AF* is common to the two triangles *DAF*, *EAF*,

the two sides *DA*, *AF* are equal to the two sides *EA*, *AF*, each to each;

and the base *DF* is equal to the base *EF*; [*Definition* 24.

therefore the angle *DAF* is equal to the angle *EAF*. [I. 8.

Wherefore *the given rectilineal angle BAC is bisected by the straight line AF*. q.e.f.

PROPOSITION 10. *PROBLEM.*

To bisect a given finite straight line, that is to divide it into two equal parts.

Let*AB*be the given straight line; it is required to divide it into two equal parts.

Describe on it an equilateral triangle *ABC*, [I. 1.

and bisect the angle *ACB* by the straight line *CD*, meeting *AB* at *D*. [I. 9.

*AB* shall be cut into two equal parts at the point *D*.

Because *AC* is equal to *CB*, [*Definition* 24.

and *CD* is common to the two triangles *ACD*, *BCD*,

the two sides *AC*, *CD* are equal to the two sides *BC*, *CD*, each to each;

and the angle *ACD* is equal to the angle *BCD*; [*Constr*.

therefore the base *AD* is equal to the base *DB*. [I. 4.

Wherefore *the given straight line AB is divided into two equal parts at the point D*. q.e.f.

PROPOSITION 11. *PROBLEM*.

To draw a straight line at right angles to a given straight line, from a given point in the same.

Let*AB*be the given straight line, and

*C*the given point in it: it is required to draw from the point

*C*a straight line at right angles to

*AB*.

Take any point *D* in *AC* and make *CE* equal to *CD* [I. 3

On *DE* describe the equilateral triangle *DFE*, [I. 1

and join *CF*.

The straight line *CF* drawn from the given point *C* shall be at right angles to the given straight line *AB*.

Because *DC* is equal to *CE*, [*Construction*.

and *CF* is common to the two triangles *DCF*, *ECF*;

the two sides *DC*, *CF* are equal to the two sides *EC*, *CF*, each to each;

and the base *DF* is equal to the base *EF*; [*Definition* 24.

therefore the angle *DCF* is equal to the angle *ECF*; [I. 8.

and they are adjacent angles.

But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; [*Definition* 10.

therefore each of the angles *DCF*, *ECF* is a right angle.

Wherefore *from the given point C in the given straight line AB, CF has been drawn at right angles to AB*. q.e.f.

Corollary. By the help of this problem it may be shewn that two straight lines cannot have a common segment.

If it be possible, let the two straight lines*ABC*,

*ABD*have the segment

*AB*common to both of them.

From the point *B* draw *BE* at right angles to *AB*.

Then, because *ABC* is a straight line, [Hypothesis.

the angle *CBE* is equal to the angle *EBA*. [*Definition* 10.

Also, because *ABD* is a straight line, [*Hypothesis*.

the angle *DBE* is equal to the angle *EBA*.

Therefore the angle *DBE* is equal to the angle *CBE*, [*Ax*. 1.

the less to the greater; which is impossible. [*Axiom* 9.

Wherefore *two straight lines cannot have a common segment*.

PROPOSITION 12. *PROBLEM*.

To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.

Let *AB* be the given straight line, which may be produced to any length both ways, and let *C* be the given point without it: it is required to draw from the point *C* a straight line perpendicular to *AB*.

*D*on the other side of

*AB*, and from the centre

*C*, at the distance

*CD*, describe the circle

*EGF*, meeting

*AB*at

*F*and

*G*. [

*Postulate*3.

Bisect *FG* at *H*, [I. 10.

and join *CH*.

The straight line *CH* drawn from the given point *C* shall be perpendicular to the given straight line *AB*.

Join *CF*, *CG*.

Because *FH* is equal to *HG*, [*Construction.*

and *HC* is common to the two triangles *FHC*, *GHC*;

the two sides *FH*,*HC* are equal to the two sides *GH*, *HC*, each to each;

and the base *CF* is equal to the base *CG*; [*Definition* 15.

therefore the angle *CHF* is equal to the angle *CHG*; [I. 8.

and they are adjacent angles.

But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it. [*Def.* 10.

Wherefore *a perpendicular CH has been drawn to the given straight line AB from the given point C without it.* q.e.f.

PROPOSITION 13. *THEOREM*.

The angles which one straight line makes with another straight line on one side of it, either are two right angles, or are together equal to two right angles.

Let the straight line *AB* make with the straight line *CD*, on one side of it, the angles *CBA*, *ABD*: these either are two right angles, or are together equal to two right angles.

For if the angle *CBA* is equal to the angle *ABD*, each of them is a right angle. [*Definition* 10.

But if not, from the point *B* draw *BE* at right angles to *CD*; [I. 11.

therefore the angles *CBE*, *EBD* are two right angles, [*Def*. 10.

Now the angle *CBE* is equal to the two angles *CBA*, *ABE*; to each of these equals add the angle *EBD*;

therefore the angles *CBE*, *EBD* are equal to the three angles *CBA*, *ABE*, *EBD*. [*Axiom* 2.

Again, the angle *DBA* is equal to the two angles *DBE*, *EBA*;

to each of these equals add the angle *ABC*;

therefore the angles *DBA*, *ABC* are equal to the three angles *DBE*, *EBA*, *ABC*. [*Axiom* 2.

But the angles *CBE*, *EBD* have been shewn to be equal to the same three angles.

Therefore the angles *CBE*, *EBD* are equal to the angles *DBA*, *ABC* [*Axiom* 1.

But *CBE*, *EBD* are two right angles;

therefore *DBA*, *ABC* are together equal to two right angles.

Wherefore, *the angles* &c. q.e.d.

PROPOSITION 14. *THEOREM*.

If, at a point in a straight line, two other straight lines, on the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

At the point *B* in the straight line *AB*, let the two straight lines *BC*, *BD*, on the opposite sides of *AB*, make the adjacent angles *ABC*, *ABD* together equal to two right angles: *BD* shall be in the same straight line with *CB*.

*BD*be not in a the same straight line with

*CB*, let

*BE*be in the same straight line with it.

Then because the straight line *AB* makes with the straight line *CBE*, on one side of it, the angles *ABC*, *ABE*, these angles are together equal to two right angles. [I. 13.

But the angles *ABC*, *ABD* are also together equal to two right angles. [*Hypothesis*.

Therefore the angles *ABC*, *ABE* are equal to the angles *ABC*, *ABD*.

From each of these equals take away the common angle *ABC*, and the remaining angle ABE is equal to the remaining angle *ABD*, [*Axiom* 3.

the less to the greater; which is impossible.

Therefore *BE* is not in the same straight line with *CB*.

And in the same manner it may be shewn that no other can be in the same straight line with it but *BD*;

therefore *BD* is in the same straight line with *CB*.

Wherefore, *if at a point* &c. q.e.d.

PROPOSITION 15. *THEOREM*.

If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

Let the two straight lines *AB*, *CD* cut one another at the point *E*; the angle *AEC* shall be equal to the angle *DEB*, and the angle *CEB*

*AED*.

Because the straight line *AE* makes with the straight line *CD* the angles *CEA*, *AED*, those angles are together equal to two right angles. [I. 13.

Again, because the straight line *DE* makes with the straight line *AB* the angles *AED*, *DEB*, these also are together equal to two right angles. [I. 13.

But the angles *CEA*, *AED* have been shewn to be together equal to two right angles.

Therefore the angles *CEA*,*AED* are equal to the angles *AED*, *DEB*.

From each of these equals take away the common angle *AED*, and the remaining angle *CEA* is equal to the remaining angle DEB. [*Axiom* 3.

In the same manner it may be shewn that the angle *CEB* is equal to the angle *AED*.

Wherefore, *if two straight lines* &c. q.e.d.

Corollary 1. From this it is manifest that, if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles.

Corollary 2. And consequently, that all the angles made by any number of straight lines meeting at one point, are together equal to four right angles.

PROPOSITION 16. *THEOREM*.

If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles.

Let *ABC* be a triangle, and let one side *BC* be produced to *D*: the exterior angle *ACD* shall be greater than either of the interior opposite angles *CBA*, *BAC*.

Bisect *AC* at *E*, [I. 10.

join *BE* and produce it to *F*, making *EF* equal to *EB*, [I. 3.

and join *FC*.

Because *AE* is equal to *EC*, and *BE* to *EF*; [*Constr*.

the two sides *AE*, *EB* are equal to the two sides *CE*, *EF*, each to each;

*AEB*is equal to the angle

*CEF*, because they are opposite vertical angles; [I. 15.

therefore the triangle *AEB* is equal to the triangle *CEF*, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; [I. 4.

therefore the angle *BAE* is equal to the angle *ECF*.

But the angle *ECD* is greater than the angle *ECF*. [*Axiom* 9.

Therefore the angle *ACD* is greater than the angle *BAE*.

In the same manner if *BC* be bisected, and the side *AC* be produced to *G*, it may be shewn that the angle *BCG*, that is the angle *ACD*, is greater than the angle *ABC*. [I. 15.

Wherefore, *if one side* &c. q.e.d.

PROPOSITION 17. *THEOREM*.

Any two angles of a triangle are together less than two right angles.

Let *ABC* be a triangle: any two of its angles are together less than two right angles.

Produce *BC* to *D*.

Then because *ACD* is the exterior angle of the triangle *ABC*, it is greater than the interior opposite angle *ABC*. [I. 16.

To each of these add the angle *ACB*

Therefore the angles *ACD*, *ACB* are greater than the angles *ABC* *ACB*.

But the angles ACD, ACB are together equal to two right angles. [I. 13.

Therefore the angles *ABC*, *ACB* are together less than two right angles.

In the same manner it may be shewn that the angles *BAC*, *ACB*, as also the angles *CAB*, *ABC*, are together less than two right angles.

Wherefore, *any two angles* &c. q.e.d.

PROPOSITION 18. *THEOREM*.

The greater side of every triangle has the greater angle opposite to it.

Let *ABC* be a triangle, of which the side *AC* is greater than the side *AB*: the angle *ABC* is also greater than the angle *ACB*.

Because *AC* is greater than

*AB*, make

*AD*equal to

*AB*, [I. 3.

and join *BD*.

Then, because *ADB* is the exterior angle of the triangle *BDC*, it is greater than the interior opposite angle *DCB*. [I. 16.

But the angle *ADB* is equal to the angle *ABD*, [I. 5.

because the side *AD* is equal to the side *AB*. [*Constr*.

Therefore the angle *ABD* is also greater than the angle *ACB*.

Much more then is the angle *ABC* greater than the angle *ACB*. [*Axiom* 9.

Wherefore, *the greater side* &c. q.e.d.

PROPOSITION 19. *THEOREM*.

The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

Let *ABC* be a triangle, of which the angle *ABC* is greater than the angle *ACB*: the side *AC* is also greater than the side *AB*.

For if not, *AC* must be either

*AB*or less than

*AB*.

But *AC* is not equal to *AB*,

for then the angle *ABC* would be equal to the angle *ACB*; [I. 5.

but it is not; [*Hypothesis*.

therefore *AC* is not equal to *AB*. Neither is *AC* less than *AB*,

for then the angle *ABC* would be less than the angle *ACB*; [I. 18.
but it is not; [*Hypothesis*.

therefore *AC* is not less than *AB*.

And it has been shewn that *AC* is not equal to *AB*.

Therefore *AC* is greater than *AB*.

Wherefore, the *greater angle* &c. q.e.d.

PROPOSITION 20. *THEOREM*.

Any two sides of a triangle are together greater than the third side.

Let *ABC* be a triangle: any two sides of it are together greater than the third side;

*BA*,

*AC*greater than

*BC*; and

*AB*,

*BC*greater than

*AC*; and

*BC*,

*CA*greater than

*AB*.

Produce *BA* to *D*,

making *AD* equal to *AC*, [I. 3.

and join *DC*.

Then, because *AD* is equal to *AC*, [*Construction*.

the angle *ADC* is equal to the angle *ACD*. [I. 5.

But the angle *BCD* is greater than the angle *ACD*. [*Ax*. 9.

Therefore the angle *BCD* is greater than the angle *BDC*.

And because the angle *BCD* of the triangle *BCD* is greater than its angle *BDC*, and that the greater angle is subtended by the greater side; [I. 19.

therefore the side *BD* is greater than the side *BC*.

But *BD* is equal to *BA* and *AC*.

Therefore *BA*, *AC* are greater than *BC*.

In the same manner it may be shewn that *AB*, *BC* are greater than *AC*, and *BC*, *CA* greater than *AB*.

Wherefore, *any two sides* &c. q.e.d.

PROPOSITION 21. *THEOREM.*

If from the ends of the side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.

Let *ABC* be a triangle, and from the points *B*, *C*, the ends of the side *BC*,

*BD*,

*CD*be drawn to the point

*D*within the triangle:

*BD*,

*DC*shall be less than the other two sides

*BA*,

*AC*of the triangle, but shall contain an angle

*BDC*greater than the angle

*BAC*.

Produce *BD* to meet *AC* at E*.*

Because two sides of a triangle are greater than the third side, the two sides *BA*, *AE* of the triangle *ABE* are greater than the side *BE*. [I. 20.

To each of these add *EC*.

Therefore *BA*, *AC* are greater than *BE*, *EC*.

Again; the two sides *CE*, *ED* of the triangle *CED* are greater than the third side *CD*. [I. 20.

To each of these add *DB*.

Therefore *CE*, *EB* are greater than *CD*, *DB*.

But it has been shewn that *BA*, *AC* are greater than *BE*, *EC*;

much more then are *BA*, *AC* greater than *BD*, *DC*.

Again, because the exterior angle of any triangle is greater than the interior opposite angle, the exterior angle *BDC* of the triangle *CDE* is greater than the angle *CED*. [I. 16.

For the same reason, the exterior angle *CEB* of the triangle *ABE* is greater than the angle *BAE*.

But it has been shewn that the angle *BDC* is greater than the angle *CEB*;

much more then is the angle *BDC* greater than the angle *BAC*.

Wherefore, *if from the ends* &c. q.e.d.

PROPOSITION 22. *PROBLEM*.

To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third.

Let *A*, *B*, *C* be the three given straight lines, of which any two whatever are greater than the third; namely, *A* and *B* greater than *C*; *A* and *C* greater than *B*; and *B* and *C* greater than *A*: it is required to make a triangle of which the sides shall be equal to *A*, *B*, *C*, each to each.

*DE*terminated at the point

*D*, but unlimited towards

*E*, and make

*DF*equal to

*A*,

*FG*equal to

*B*, and

*GH*equal to

*C*. [I. 3.

From the centre *F*, at the distance *FD*, describe the circle *DKL*. [*Post.* 3.

From the centre *G*, at the distance *GH*, describe the circle *HLK*, cutting the former circle at *K*.

Join *KF*, *KG*. The triangle *KFG* shall have its sides equal to the three straight lines *A*,*B*, *C*.

Because the point *F* is the centre of the circle *DKL*, *FD* is equal to *FK*. [*Definition* 15.

But *FD* is equal to *A*. [*Construction*.

Therefore *FK* is equal to *A*. [*Axiom* 1.

Again, because the point *G* is the centre of the circle *HLK*, *GH* is equal to *GK* [*Definition* 15.

But *GH* is equal to *C*. [*Construction*.

Therefore *GK* is equal to *C*. [*Axiom* 1.

And *FG* is equal to *B*. [*Construction.*

Therefore the three straight lines *KF*, *FG*, *GK* are equal to the three *A*, *B*, *C*.

Wherefore *the triangle KFG has its three sides KF*, *FG*, *GK equal to the three given straight lines A*, *B*, *C*. q.e.f.

PROPOSITION 23. *PROBLEM*.

At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.

Let *AB* be the given straight line, and *A* the given point in it, and *DCE* the given rectilineal angle: it is required to make at the given point *A*, in the given straight line *AB*, an angle equal to the given rectilineal angle *DCE*.

In *CD*, *CE* take any points *D*, *E*, and join *DE*.

Make the triangle *AFG* the sides of which shall be equal to the three straight lines *CD*, *DE*, *EC*; so that *AF* shall be equal to *CD*, *AG* to *CE*, and *FG* to *DE*. [I. 22.

The angle *FAG* shall be equal to the angle *DCE*.

Because *FA*, *AG* are equal to *DC*, *CE*, each to each, and the base *FG* equal to the base *DE*; [*Construction*.

therefore the angle *FAG* is equal to the angle *DCE*. [I. 8.

Wherefore *at the given point A in the given straight line AB, the angle FAG has been made equal to the given rectilineal angle DCE*. q.e.f.

PROPOSITION 24. *THEOREM*.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other, the base of that which has the greater angle shall be greater than the base of the other.

Let *ABC*, *DEF* be two triangles, which have the two sides *AB*, *AC*, equal to the two sides *DE*, *DF*, each to each, namely, *AB* to *DE*, and *AC* to *DF*, but the angle *BAC* greater than the angle *EDF*: the base *BC* shall be

greater than the base *EF*.

Of the two sides *DE*, *DF*, let *DE* be the side which is not greater than the other. At the point *D* in the straight line *DE*, make the angle *EDG* equal to the angle *BAC*, [I. 23.

and make *DG* equal to *AC* or *DF*, [I. 3.

and join *EG*, *GF*.

Because *AB* is equal to *DE*, [*Hypothesis*.

and *AC* to *DG*; [*Construction*.

the two sides *BA*, *AC* are equal to the two sides *ED*, *DG*, each to each;

and the angle *BAC* is equal to the angle *EDG*; [*Constr*.

therefore the base *BC* is equal to the base *EG*. [I. 4.

And because *DG* is equal to *DF*, [*Construction*.

the angle *DGF* is equal to the angle *DFG*. [I. 5.

But the angle *DGF* is greater than the angle *EGF*. [*Ax*. 9.

Therefore the angle *DFG* is greater than the angle *EGF*. Much more then is the angle *EFG* greater than the angle *EGF*. [*Axiom* 9.

And because the angle *EFG* of the triangle *EFG* is greater than its angle *EGF*, and that the greater angle is subtended by the greater side, [I. 19.

therefore the side *EG* is greater than the side *EF*.

But *EG* was shewn to be equal to *BC*;

therefore *BC* is greater than *EF*.

Wherefore, *if two triangles* &c. q.e.d.

PROPOSITION 25. *THEOREM*.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other, the angle contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides equal to them, of the other.

Let *ABC*, *DEF* be two triangles, which have the two sides *AB*, *AC* equal to the two sides *DE*, *DF*, each to each, namely, *AB* to *DE*, and *AC* to *DF*, but the base *BC* greater than the base *EF*: the angle *BAC* shall be greater than the angle *EDF*.

For if not, the angle *BAC* must be either equal to the angle *EDF* or less than the angle *EDF*.

But the angle *BAC* is not equal to the angle *EDF*, for then the base *BC* would be equal to the base *EF*; [I. 4.

but it is not; [*Hypothesis*.

therefore the angle *BAC* is not equal to the angle *EDF*.

Neither is the angle *BAC* less than the angle *EDF*,

for then the base *BC* would be less than the base *EF*; [I. 24.

but it is not; [*Hypothesis.*

therefore the angle *BAC* is not less than the angle *EDF*.

And it has been shewn that the angle *BAC* is not equal to the angle *EDF*.

Therefore the angle *BAC* is greater than the angle *EDF*.

Wherefore, *if two triangles* &c. q.e.d.

PROPOSITION 26. *THEOREM*.

If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, either the sides adjacent to the equal angles, or sides which are opposite to equal angles in each, then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other.

Let *ABC*, *DEF* be two triangles, which have the angles *ABC*, *BCA* equal to the angles *DEF*, *EFD*, each to each, namely, *ABC* to *DEF*, and *BCA* to *EFD*; and let them have also one side equal to one side; and first let those sides be equal which are adjacent to the equal angles in the two triangles, namely, *BC* to *EF*: the other sides shall be equal, each to each, namely, *AB* to *DE*, and *AC* to *DF*, and the third

*BAC*equal to the third angle

*EDF*.

For if *AB* be not equal to *DE*, one of them must be greater than the other. Let *AB* be the greater, and make *BG* equal to *DE*, [I. 3.

and join *GC*.

Then because *GB* is equal to *DE*, [*Construction*.

and *BC* to *EF*; [*Hypothesis*.

the two sides *GB*, *BC* are equal to the two sides *DE*, *EF*, each to each;

and the angle *GBC* is equal to the angle *DEF*; [*Hypothesis*.

therefore the triangle *GBC* is equal to the triangle *DEF*, and the other angles to the other angles, each to each, to which the equal sides are opposite; [I. 4.

therefore the angle *GCB* is equal to the angle *DFE*.

But the angle *DFE* is equal to the angle *ACB*. [*Hypothesis*.

Therefore the angle *GCB* is equal to the angle *ACB*, [*Ax*. 1.

the less to the greater; which is impossible.

Therefore *AB* is not unequal to *DE*,

that is, it is equal to it;

and *BC* is equal to *EF*; [*Hypothesis*.

therefore the two sides *AB*, *BC* are equal to the two sides *DE*, *EF*, each to each;

and the angle *ABC* is equal to the angle *DEF*; [*Hypothesis*.

therefore the base *AC* is equal to the base *DF*, and the third angle *BAC* to the third angle *EDF*. [I. 4.

Next, let sides which are opposite to equal angles in each triangle be equal to one another, namely, *AB* to *DE*: likewise in this case the other sides shall be equal, each to each, namely, *BC* to *EF*, and *AC* to *DF*, and also the third angle *BAC* equal to the third angle *EDF*.

For if *BC* be not equal to *EF*, one of them must be greater than the other.

Let *BC* be the greater, and make *BH* equal to *EF*, [I. 3.

and join *AH*.

Then because *BH* is equal to *EF*, [*Construction.*

and *AB* to *DE*; [*Hypothesis*.

the two sides *AB*, *BH* are equal to the two sides *DE*, *EF*, each to each;

and the angle *ABH* is equal to the angle *DEF*; [*Hypothesis*.

therefore the triangle *ABH* is equal to the triangle *DEF*, and the other angles to the other angles, each to each, to which the equal sides are opposite; [I. 4.

therefore the angle *BHA* is equal to the angle *EFD*.

But the angle *EFD* is equal to the angle *BCA*. [*Hypothesis*.

Therefore the angle *BHA* is equal to the angle *BCA*; [*Ax*. 1.

that is, the exterior angle *BHA* of the triangle *AHC* is equal to its interior opposite angle *BCA*;

which is impossible. [I. 16.

Therefore *BC* is not unequal to *EF*,

that is, it is equal to it;

and *AB* is equal to *DE*; [*Hypothesis*.

therefore the two sides *AB*, *BC* are equal to the two sides *DE*, *EF*, each to each;

and the angle *ABC* is equal to the angle *DEF*; [Hypothesis.

therefore the base *AC* is equal to the base *DF*, and the third angle *BAC* to the third angle *EDF*. [I. 4.

Wherefore, *if two triangles* &c. q.e.d.

*THEOREM*.

*If a straight line falling on two other straight lines, make the alternate angles equal to one another, the two straight lines shall be parallel to one another.*

Let the straight line *EF*, which falls on the two straight
lines *AB*, *CD*, make the alternate angles *AEF*, *EFD*
equal to one another: *AB* shall be parallel to *CD*.

For if not, *AB* and *CD*, being produced, will meet
either towards *B*, *D* or towards *A*, *C*. Let them be pro-
duced and meet towards *B*, *D* at the point *G*.

Therefore *GEF* is a triangle, and its exterior angle *AEF*
is greater than the interior opposite angle *EFG*; [I. 16.

But the angle *AEF* is also equal to the angle *EFG*; [*Hyp*.

which is impossible.

Therefore *AB* and *CD* being produced, do not meet to-
wards *B*, *D*.

In the same manner, it may be shewn that they do not
meet towards *A*, *C*.

But those straight lines which being produced ever so far
both ways do not meet, are parallel. [*Definition* 35.

Therefore *AB* is parallel to *CD*.

Wherefore, *if a straight line* &c. q.e.d.

*THEOREM*.

*If a straight line falling on two other straight lines, make the exterior angle equal to the interior and opposite angle on the same side of the line, or make the interior angles on the same side together equal to two right angles, the two straight lines shall be parallel to one another.* Let the straight line *EF*, which falls on the two
straight lines *AB*, *CD*, make the exterior angle *EGB*
equal to the interior and opposite angle *GHD* on the same
side, or make the interior angles on the same side *BGH*,
*GHD* together equal to two right angles : *AB* shall be
parallel to *CD*.

Because the angle *EGB* is
equal to the angle GHD, [I. 15.

and the angle EGB is also equal
to the angle *AGH*, [1.15.

therefore the angle *AGH* is
equal to the angle *GHD*;[*Ax*.l.

and they are alternate angles ;

therefore *AB* is parallel to
*CD*. [I. 27.

Again; because the angles *BGH*, *GHD* are together
equal to two right angles, [*Hypothesis*.

and the angles *AGH*, *BGH* are also together equal to two
right angles, [1. 13.

therefore the angles *AGH*, *BGH* are equal to the angles
*BGH*, *GHD*.
Takeaway the common angle *BGH*; therefore the remaining
angle *AGH* is equal to the remaining angle *GHD*; [*Axiom* 3.
and they are alternate angles ;
therefore *AB* is parallel to *CD*. [I. 27.

Wherefore, *if a straight line* &c. q.e.d.

*THEOREM*.

*If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles*.

Let the straight line *EF* fall on the two parallel
straight lines *AB*, *CD* : the alternate angles *AGH*, *GHD*
shall be equal to one another, and the exterior angle
*EGB* shall be equal to the interior and opposite angle on the same side, *GHD*, and the two interior angles on
the same side, *BGH*, *GHD*, shall be together equal to two
right angles.

For if the angle *AGH* be
not equal to the angle *GHD*,
one of them must be greater
than the other ; let the angle
*AGH* be the greater.

Then the angle *AGH* is greater
than the angle *GHD* ;

to each of them add the angle
*BGH*;

therefore the angles *AGH*, *BGH* are greater than the
angles *BGH*, *GHD*.

But the angles *AGH*, *BGH* are together equal to two
right angles ; [I. 13.

therefore the angles *BGH*, *GHD* are together less than
two right angles.

But if a straight line meet two straight lines, so as to
make the two interior angles on the same side of it, taken
together, less than two right angles, these straight lines
being continually produced, shall at length meet on that
side on which are the angles which are less than two
right angles. [*Axiom* 12.

Therefore the straight lines AB, CD, if continually pro-
duced, will meet.

But they never meet, since they are parallel by hypothesis.

Therefore the angle *AGH* is not unequal to the angle
*GHD* ; that is, it is equal to it.

But the angle *AGH* is equal to the angle *EGB*. [I. 15.

Therefore the angle *EGB* is equal to the angle *GHD*. [*Ax*. 1 .

Add to each of these the angle *BGH*.

Therefore the angles *EGB*, *BGH* are equal to the angles
*BGH*, *GHD*. [*Axiom* 2.

But the angles *EGB*, *BGH* are together equal to two
right angles. [I. 13.

Therefore the angles *BGH*, *GHD* are together equal to
two right angles. [*Axiom* 1.

*if a straight line*&c. q.e.d.

*THEOREM*.

*Straight lines which are parallel to the same straight line are parallel to each other*.

Let *AB*, *CD* be each of them parallel to *EF*: *AB*
shall be parallel to *CD*.

Let the straight line *GHK*
cut *AB*, *EF*, *CD*.

Then, because *GHK* cuts
the parallel straight lines *AB*,
*EF*, the angle *AGH* is equal
to the angle *GHF*. [I. 29.

Again, because *GK* cuts
the parallel straight lines *EF*,
*CD*, the angle *GHF* is equal
to the angle *GKD*. [I. 29.

And it was shewn that the
angle *AGK* is equal to the angle *GHF*.

Therefore the angle *AGK* is equal to the angle *GKD*;[*Ax*. 1.
and they are alternate angles ;

therefore *AB* is parallel to *CD*.

Wherefore, *straight lines* &c. q.e.d.

*PROBLEM*.

*To draw a straight line through a given point parallel to a given straight line*.

Let *A* be the given point, and *BC* the given straight
line : it is required to draw a straight line through the
point *A* parallel to the straight line *BC*.
In *BC* take any point
*D*, and join *AD* ; at the
point *A* in the straight
line *AD*, make the angle
*DAE* equal to the angle
*ADC*, [I.23.

and produce the straight line *EA* to *F*.
*EF* shall be parallel to *BC*. Because the straight line *AD* which meets the two
straight lines *BC*, *EF*, makes the alternate angles *BAD*,
*ADC* equal to one another, [*Construction*.*EF* is parallel to *BC*. [I. 27.

Wherefore the straight line *EAF* is drawn through the
given point *A*, parallel to the given straight line *BC*. q.e.f.

*THEOREM*.

*If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the*
three interior angles of every triangle are toge-
ther equal to two right angles*. *

Let *ABC* be a triangle, and let one of its sides *BC*
be produced to *D* : the exterior angle *ACD* shall be equal
to the two interior and opposite angles *CAB*,*ABC*; and
the three interior angles of the triangle, namely, *ABC*,
*BCA*, *CAB* shall be equal to two right angles.

Through the point *C* draw
*CE* parallel to *AB*. [1.31.

Then, because *AB* is par-
allel to *CE*, and *AC* falls on
them, the alternate angles
*BAC*, *ACE* are equal. [I. 29.

Again, because *AB* is parallel to *CE*, and *BD* falls on
them, the exterior angle *ECD* is equal to the interior and
opposite angle *ABC*. [I. 29.

But the angle *ACE* was shewn to be equal to the angle
*BAC*;

therefore the whole exterior angle *ACD* is equal to the
two interior and opposite angles *CAB*, *ABC*. [*Axiom* 2.

To each of these equals add the angle *ACB*;

therefore the angles *ACD*, *ACB* are equal to the three
angles *CBA*, *BAC*, *ACB*. [*Axiom* 2.

But the angles *ACD*, *ACB* are together equal to two right
angles ; [I. 13.

Therefore also the angles *CBA*, *BAC*, *ACB* are together
equal to two right angles. [*Axiom* 1.

Wherefore, *if a side of any triangle* &c. q.e.d. Corollary 1. *All the interior angles of any recti-lineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides*.

For any rectilineal figure *ABCDE* can be divided into
as many triangles as the figure has sides, by drawing;
straight lines from a point *F* within the figure to each of its angles.

And by the preceding proposition,
all the angles of these triangles are
equal to twice as many right angles
as there are triangles, that is, as the
figure has sides.

And the same angles are equal to the
interior angles of the figure, together
with the angles at the point *F*, which
is the common vertex of the triangles,
that is, together with four right angles. [I. 15. *Corollary* 2.

Therefore all the interior angles of the figure, together with
four right angles, are equal to twice as many right angles
as the figure has sides.

Corollary 2. *All the exterior angles of any recti-lineal figure are together equal to four right angles*.

Because every interior angle
*ABC*, with its adjacent exterior
angle *ABD*, is equal to two
right angles ; [I. 13.

therefore all the interior angles
of the figure, together with all
its exterior angles, are equal to
twice as many right angles as
the figure has sides.

But, by the foregoing Corollary all the interior angles of the
figure, together with four right angles, are equal to twice
as many right angles as the figure has sides.

Therefore all the interior angles of the figure, together with
all its exterior angles, are equal to all the interior angles of
the figure, together with four right angles.

Therefore all the exterior angles are equal to four right;

*THEOREM*.

*The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are also themselves equal and parallel*.

Let *AB* and *CD* be equal and parallel straight lines,
and let them be joined towards the same parts by the
straight lines *AC* and *BD* : *AC* and *BD* shall be equal
and parallel.

Then because *AB* is par-
allel to *CD*, [*Hypothesis*.

and *BC* meets them,
the alternate angles *ABC*,
*BCD* are equal. [I. 29.

And because *AB* is equal to *CD*, [*Hypothesis*.
and *BC* is common to the two triangles *ABC*, *DCB* ;

the two sides *AB*, *BC* are equal to the two sides *DC*, *CB*,
each to each ;

and the angle *ABC* was shewn to be equal to the angle
*BCD*;

therefore the base *AC* is equal to the base *BD*, and the
triangle *ABC* to the triangle *BCD*, and the other angles
to the other angles, each to each, to which the equal sides
are opposite ; [I. 4.

therefore the angle *ACB* is equal to the angle *CBD*.

And because the straight line *BC* meets the two straight
lines *AC*, *BD*, and makes the alternate angles ACB, *CBD*
equal to one another, *AC* is parallel to *BD*, [I. 27.

And it was shewn to be equal to it.

Wherefore, *the straight lines* &c. q.e.d.

*THEOREM*.

*The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects the par- allelogram, that is, divides it into two equal parts*.

*Note*. A parallelogram is a four-sided figure of which the
opposite sides are parallel ; and a diameter is the straight line
joining two of its opposite angles. Let *ACDB* be a parallelogram, of which *BC* is a diameter; the opposite sides and angles of the figure shall be equal to one another, and the diameter *BC* shall bisect it.

Because *AB* is parallel to *CD*, and *BC* meets them, the alternate angles *ABC*, *BCD* are equal to one another. [I. 29.

And because *AC* is parallel to *BD*, and *BC* meets them, the alternate angles ACB, CBD are equal to one another. [I. 29.

Therefore the two triangles *ABC*, *BCD* have two angles *ABC*, *BCA* in the one, equal to two angles *DCB*, *CBD* in the other, each to each, and one side *BC* is common to the two triangles, which is adjacent to their equal angles;

therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, namely, the side *AB* equal to the side *CD*, and the side *AC* equal to the side *BD*, and the angle *BAC* equal to the angle *CDB*. [I. 26.

And because the angle *ABC* is equal to the angle *BCD* and the angle *CBD* to the angle *ACB*, the whole angle *ABD* is equal to the whole angle *ACD*. [*Ax*. 2.

And the angle *BAC* has been shewn to be equal to the angle *CDB*.

Therefore the opposite sides and angles of a parallelogram are equal to one another.

Also the diameter bisects the parallelogram.

For *AB* being equal to *CD*, and *BC* common,

the two sides *AB*, *BC* are equal to the two sides *DC*, *CB* each to each;

and the angle *ABC* has been shewn to be equal to the angle *BCD*;

therefore the triangle *ABC* is equal to the triangle *BCD*, [1. 4.

and the diameter *BC* divides the parallelogram *ACDB* into two equal parts.

*the opposite sides*&c. q.e.d.

*THEOREM*.

*Parallelograms on the same base, and between the same parallels, are equal to one another*.

Let the parallelograms *ABCD*, *EBCF* be on the same
base *BC*, and between the same parallels *AF*, *BC* : the paral-
lelogram *ABCD* shall be equal to the parallelogram *EBCF*.

If the sides *AD*, *EF* of
the parallelograms *ABCD*,
*EBCF*, opposite to the base
*BC*, be terminated at the same
point *D*, it is plain that each of
the parallelograms is double of
the triangle *BDC*; [I. 34.

and they are therefore equal to one another. [Axiom 6.

But if the sides *AD*, *EF*, opposite to the base *BC*
of the parallelo-
grams *ABCD*,
*EBCF* be not
terminated at
the same point,
then, because
*ABCD* is a par-
allelogram *AD* is equal to *BC* ; [I. 34.

for the same reason *EF* is, equal to *BC* ;

therefore *AD* is equal to *EF*; [*Axiom* 1.

therefore the whole, or the remainder, *AE* is equal to the
whole, or the remainder, *DF*. [*Axioms* 2, 3.

And *AB* is equal to *DC* ; [I. 34.

therefore the two sides *EA*, *AB* are equal to the two sides *FD*,*DC* each to each;

and the exterior angle *FDC* is equal to the interior and
opposite angle *EAB* ; [I. 29.

therefore the triangle EAB is equal to the triangle
*FDC*. [I. 4.

Take the triangle FDC from the trapezium ABCF,
and from the same trapezium take the triangle EAB,
and the remainders are equal ; [*Axiom* 3.

that is, the parallelogram *ABCD* is equal to the parallelo-gram *EBCF*.

*parallelograms on the same base*&c. q.e.d.

*THEOREM*.

*Parallelograms on equal bases, and between the same *
parallels are equal to one another*. *

Let *ABCD*, *EFGH* be parallelograms on equal bases
*BC*, *FG* and between the same parallels *AH*, *BG*: the
parallelogram *ABCD* shall be equal to the parallelogram
*EFGH*.

Then, because *BC*
is equal to *FG*, [*Hyp*.

and *FG* to *EH*,[I.34.*BC* is equal to
*EH*; [Axiom 1.

and they are parallels, - [Hypothesis.

and joined towards the same parts by the straight lines
*BE*, *CH*.

But straight lines which join the extremities of equal and
parallel straight lines towards the same parts are them-
selves equal and parallel. [I. 33.

Therefore *BE*, *CH* are both equal and parallel.
Therefore *EBCH* is a parallelogram. [*Definition*.

And it is equal to *ABCD*, because they are on the same
base *BC*, and between the same parallels *BC*,*AH*. [I. 35.

For the same reason the parallelogram *EFGH* is equal
to the same *EBCH*.

Therefore the parallelogram *ABCD* is equal to the par-
allelogram *EFGH*. [*Axiom* l.

Wherefore, *parallelograms* &c. q.e.d.

*Triangles on the same base, and between the same par-*
allels, are equal*. *

Let the triangles *ABC*,
*DBC* be on the same base
*BC*, and between tie same
parallels *AD*,*BC* the tri-
angle *ABC* shall be equal
to the triangle *DBC*.

Produce *AD* both ways
to the points *E*, *F* ; [*Post*. 2.

through *B* draw *BE* parallel to *CA*, and through *C* draw
*CF* parallel to *BD*. [I. 31.

Then each of the figures *EBCA*, *DBCF* is a parallelo-
gram ; [*Definition*.
and *EBCA* is equal to *DBCF*, because they are on the same base *BC*, and between the same parallels *BG*, *EF*. [I. 35.

And the triangle *ABC* is, half of the parallelogram *EBCA*,
because the diameter *AB* bisects the parallelogram ; [I. 34.

and the triangle *DBC* is half of the parallelogram *DBCF*,
because the diameter *DC* bisects the parallelogram. [I. 34.

But the halves of equal things are equal. [*Axiom* 7.

Therefore the triangle *ABC* is equal to the triangle *DBC*.
Wherefore, *triangles* &c. q.e.d.

*THEOREM*.

*Triangles on equal bases, and between the same par-*
allels, are equal to one another*. *

Let the triangles *ABC*, *DEF* be on equal bases *BC*,
*EF*, and between the same parallels *BF*, *AD* : the triangle
*ABC* shall be equal to the triangle *DEF*,

Produce *AD* both
ways to the points
*G*,*H*;

through *B* draw *BG*
parallel to *CA*, and
through *F* draw *FH*
parallel to *ED*.[I.31.

Then each of the
figures *GBCA*, *DEFH* is a parallelogram. [*Definition*.

And they are equal to one another because they are on
equal bases *BC*, *EF*, and between the same parallels
*BF*, *GH*. [1. 36.

And the triangle *ABC* is half of the parallelogram *GBCA*,
because the diameter *AB* bisects the parallelogram ;[I. 34.

and the triangle *DEF* is half of the parallelogram *DEFH*,
because the diameter *DF* bisects the parallelogram.

But the halves of equal things are equal. [*Axiom* 7.

Therefore the triangle *ABC* is equal to the triangle *DEF*.

*triangles*&c. q.e.d.

*THEOREM*.

*Equal triangles on the same base, and on the same side of it, are between the same parallels*.

Let the equal triangles *ABC*, *DBC* be on the same base BC, and on the same side of it : they shall be be-
tween the same parallels.

Join *AD*. *AD* shall be parallel to *BC*.

For if it is not, through *A* draw
*AE* parallel to *BC*, meeting *BD*
at *E*. [I. 31.

and join *EC*.

Then the triangle *ABC* is equal to the triangle *EBC*,
because they are on the same base *BC*, and between the
same parallels *BC*, *AE*. [I. 37.

But the triangle *ABC* is equal to the triangle *DBC*.[*Hyp*.
Therefore also the triangle *DBC* is equal to the triangle *EBC*, [*Axiom* 1.

the greater to the less ; which is impossible.

Therefore *AE* is not parallel to *BC*.

In the same manner it can be shewn, that no other
straight line through *A* but *AD* is parallel to *BC*;

therefore *AD* is parallel to *BC*.

Wherefore, *equal triangles* &c., q.e.d.

*THEOREM*.

*Equal triangles, on equal bases, in the same straight line, and on the same side of it, are between the same parallels*.

Let the equal triangles *ABC*, *DEF* be on equal bases
*BC*, *EF*, in the same straight line *BF*, and on the same
side of it: they shall be between the same parallels.

Join *AD*.*AD* shall be parallel to *BF*.

For if it is not, through *A*
draw *AG* parallel to *BF*,

meeting *ED* at *G* [I. 31.

and join *GF*.

Then the triangle *ABC* is equal to the triangle *GEF*,
because they are on equal bases *BC*, *EF*, and between
the same parallels, [I. 38.

But the triangle ABC is equal to the triangle DEF. [*Hyp*.

Therefore also the triangle *DEF* is equal to the triangle
*GEF*, [*Axiom* 1.

the greater to the less ; which is impossible.

Therefore *AD* is not parallel to *BF*,

In the same manner it can be shewn that no other
straight line through *A* but *AD* parallel to *BF*;
therefore *AD* is parallel to *BF*,

Wherefore, *equal triangles* &c. q.e.d.

*THEOREM*.

*If a parallelogram and a triangle he on the same base and between the same parallels, the parallelogram, shall be double of the triangle*.

Let the parallelogram *ABCD* and the triangle *EBC* be
on the same base *BC*, and between the same parallels
*BC*, *AE* : the parallelogram *ABCD* shall be double of the triangle *EBC*.

Join *AC*.

Then the triangle *ABC*
is equal to the triangle *EBC*
because they are on the same
base *BC*, and between the same
parallels *BC*, *AE*. [I. 37.

But the parallelogram *ABCD*
is double of the triangle *ABC*,

because the diameter *AC* bisects the parallelogram. [I . 34.

Therefore the parallelogram *ABCD* is also double of the
triangle *EBC*.

*if a parallelogram*&c. q.e.d.

*To describe a parallelogram that shall be equal to a given triangle and have one of its angles equal to a given rectilineal angle*.

Let *ABC* be the given triangle, and *D* the given recti-
lineal angle : it is required to describe a parallelogram that
shall be equal to the given triangle *ABC*, and have one of
its angles equal to *D*.

Bisect *BC* at *E*:[I.10.

join *AE*, and at the point
*E*, in the straight line *EC*,
make the angle *CEF* equal
to *D*; [I.23.

through *A* draw *AFG*
parallel to *EC*, and through
*C* draw *CG* parallel to
*EF*. [I. 31.

Therefore *FECG* is a parallelogram. [*Definition*.

And, because *BE* is equal to *EC*, [*Construction*.
the triangle *ABE* is equal to the triangle *AEC*, because
they are on equal bases *BE*, *EC*, and between the same
parallels *BC*, *AG*. [I. 38.

Therefore the triangle *ABC* is double of the triangle *AEC*.

But the parallelogram *FECG* is also double of the triangle
*AEC*, because they are on the same base *EC*, and between
the same parallels *EC*, *AG*. [I. 41.

Therefore the parallelogram *FECG* is equal to the triangle
*ABC* ; [*Axiom* 6.

and it has one of its angles *CEF* equal to the given angle
*D*. [*Construction*.

*a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D*. q.e.f.

*THEOREM*.

*The complements of the parallelograms which are about the diameter of any parallelogram are equal to one another*.

Let *ABCD* be a parallelogram, of which the diameter
is *AC*; and *EH*, *GF* parallelograms about *AC*, that is,
through which *AC* passes ; and *BK*, *KD* the other paral-
lelograms which make up the whole figure *ABCD*, and
which are therefore called the complements: the comple-
ment *BK* shall be equal to the complement *KD*.

Because *ABCD* is a
parallelogram, and *AC* its
diameter, the triangle *ABC*
is equal to the triangle
*ADC*. [I. 34.

Again, because *AEKH* is
a parallelogram, and *AK*
its diameter, the triangle
*AEK* is equal to the triangle
*AHK*. [I. 34.

For the same reason the triangle *KGC* is equal to the
triangle *KFC*.
Therefore, because the triangle *AEK* is equal to the tri-
angle *AHK*, and the triangle *KGC* to the triangle *KFC* ;

the triangle *AEK* together with the triangle *KGC* is equal
to the triangle *AHK* together with the triangle *KFC*. [*Ax*. 2.
But the whole triangle *ABC* was shewn to be equal to the
whole triangle *ADC*.

Therefore the remainder, the complement *BK*, is equal to
the remainder, the complement *KD*. [*Axiom* 3,

Wherefore, *the complements* &c. q.e.d.

*PROBLEM*.

*To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle*. Let *AB* be the given straight line, and *C* the given
triangle, and *D* the given rectilineal angle : it is required
to apply to the straight line *AB a parallelogram equal to *
the triangle *C*, and having an angle equal to *D*.

Make the parallelogram *BEFG* equal to the triangle
*C*, and having the angle *EBG* equal to the angle *D*, so
that *BE* may be in the same straight line with *AB* ; [I. 42.

produce *FG* to *H* ;

through *A* draw *AH* parallel to *BG* or *EF*, [I. 31.

and join HB. [I. 31.

Then, because the straight line *HF* falls on the parallels
*AH*, *EF*, the angles *AHF*, *HFE* are together equal to two right angles. [I. 29.

Therefore the angles *BHF*, *HFE* are together less than two right angles.

But straight lines which with another straight line make the interior angles on the same side together less than two right angles will meet on that side, if produced far enough. [*Ax*. 12.

Therefore *HB* and *FE* will meet if produced ;

let them meet at *K*.

Through *K* draw *KL* parallel to *EA* or *FH* ; [I. 31.

and produce *HA*, *GB* to the points *L*, *M*.

Then *HLKF* is a parallelogram, of which the diameter
is *HK*; and *AG*, *ME* are parallelograms about *HK*; and
*LB*, *BF* are the complements.

Therefore *LB* is equal to *BF*. [I. 43.

But *BF* is equal to the triangle *C* [*Construction*.</br.>
Therefore *LB* is equal to the triangle *C* [*Axiom* 1.

And because the angle GBE is equal to the angle *ABM*, [I.15.

and likewise to the angle *D* ; [*Construction*.

the angle *ABM* is equal to the angle *D*. [*Axiom* 1.

Wherefore *to the given straight line AB the parallelo-*
gram LB is applied, equal to the triangle *C*, and having the angle *ABM* equal to the angle *D*. q.e.f.

*PROBLEM*.

*To describe a parallelogram equal to a given rectilineal *
figure, and having an angle equal to a given rectilineal
angle*. *

Let *ABCD* be the given rectilineal figure, and *E* the
given rectilineal angle: it is required to describe a par-
allelogram equal to *ABCD*, and having an angle equal to *E*.

Join *DB*, and describe the parallelogram *FH* equal to
the triangle *ADB*, and having the angle *FKH* equal to the angle *E* ; [I. 42.

and to the straight line *GH* apply the parallelogram *GM*
equal to the triangle *DBC*, and having the angle *GHM*
equal to the angle *E*. [I. 44.

The figure *FKML* shall be the parallelogram required.
Because the angle *E* is equal to each of the angles *FKH*, *GHM*, [*Construction*.

the angle *FKH* is, equal to the angle *GHM*. [*Axiom* 1.

Add to each of these equals the angle *KHG* ;

therefore the angles *FKH*, *KHG* are equal to the angles *KHG*,*GHM*. [*Axiom* 2.

But *FKH*, *KHG* are together equal to two right angles;[I.29.

therefore *KHG*,*GHM* are together equal to two right angles.

And because at the point *H* in the straight line *GH*, the
two straight lines *KH*, *HM*, on the opposite sides of it,
make the adjacent angles together equal to two right angles,
*KH* is in the same straight line with *HM*. [I. 14.

And because the straight line *HG* meets the parallels
*KM*, *FG*, the alternate angles *MHG*, *HGF* are equal. [I. 29.

Add to each of these equals the angle *HGL* ;

therefore the angles *MHG*, *HGL*, are equal to the angles *HGF*, *HGL*. [*Axiom* 2.

But *MHG*,*HGL* are together equal to two right angles; [I. 29.

therefore *HGF*, *HGL* are together equal to two right angles.
Therefore *FG* is in the same straight line with *GL*. [I. 14.

And because *KF* is parallel to *HG*, and *HG* to *ML*,[*Constr*.*KF* is parallel to *ML* ; [I. 30.

and *KM*, *FL* are parallels ; [*Construction*.

therefore *KFLM* a parallelogram. [*Definition*.

And because the triangle *ABD* is equal to the parallelo-
gram *HF*, [*Construction*.

and the triangle *DBC* to the parallelogram *GM* ; [*Constr*.
the whole rectilineal figure *ABCD* is equal to the whole
parallelogram *KFLM*. [*Axiom* 2.

Wherefore, the parallelogram *KFLM* has been de-
scribed equal to the given rectilineal figure *ABCD*, and
having the angle *FKM* equal to the given angle *E*. q.e.f.

Corollary. From this it is manifest, how to a given
straight line, to apply a parallelogram, which shall have an
angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure ; namely, by applying to the
given straight line a parallelogram equal to the first tri-
angle *ABD*, and having an angle equal to the given angle;

*PROBLEM*.

*To describe a square on a given straight line*.

Let *AB* be the given straight line : it is required to describe a square on *AB*.

From the point *A* draw *AC*
at right angles to *AB*; [I. 11.

and make *AD* equal to *AB*; [I. 3.

through *D* draw *DE* parallel to
*AB* ; and through B draw *BE*
parallel to *AD*. [I. 31.*ADEB* shall be a square.

For *ADEB* is by construction
a parallelogram ;

therefore *AB* is equal to *DE*
and *AD* to *BE*. [I. 34.

But *AB* is equal to *AD*. [*Construction*.
Therefore the four straight lines *BA*,*AD*, *DE*, *EB* are equal to one another, and the parallelogram *ADEB* is
equilateral. [*Axiom* 1.

Likewise all its angles are right angles.

For since the straight line *AD* meets the parallels *AB*,*DE*, the angles *BAD*, *ADE* are together equal to two
right angles ; [I. 29.

but *BAD* is a right angle ; [*Construction*.

therefore also *ADE* is a right angle. [Axiom 3.

But the opposite angles of parallelograms are equal. [I. 34.

Therefore each of the opposite angles *ABE*, *BED* is a
right angle. [*Axiom* 1.

Therefore the figure *ADEB* is rectangular;
and it has been shewn to be equilateral.
Therefore *it is a square*. [Definition, 30.
*And it is described on the given straight line AB*. q.e.f.

Corollary. From the demonstration it is manifest that every parallelogram which has one right angle has all its

angles right angles. *THEOREM*.

*In any right-angled triangle, the square which is de-scribed on the side subtending the right angle is equal to the squares described on the sides which contain the right angle*.

Let *ABC* be a right-angled triangle, having the right
angle *BAC* : the square described on the side *BC* shall be
equal to the squares described on the sides *BA*, *AC*.

On *BC* describe
the square *BDEC*,
and on *BA*, *AC* de-
scribe the squares
*GB*,*HC*; [I.46.

through *A* draw *AL*
parallel to *BD* or
*CE* ; [I. 31.

and join *AD*, *FC*.

Then, because the
angle *BAC* is a right
angle, [*Hypothesis*.

and that the angle
*BAG* is also a right
angle, [*Definition* 30.

the two straight lines *AC*, *AG*, on the opposite sides of
*AB*, make with it at the point *A* the adjacent angles equal
to two right angles ;

therefore *CA* is in the same straight line with *AG*. [I, 14.

For the same reason, *AB* and *AH* are in the same straight line.

Now the angle *DBC* is equal to the angle *FBA*, for each
of them is a right angle. [*Axiom* 11.

Add to each the angle *ABC*.

Therefore the whole angle *DBA* is equal to the whole angle
*FBC*. [*Axiom* 2.

And because the two sides *AB*, *BD* are equal to the two
sides *FB*, *BC*, each to each ; [*Definition* 30.

and the angle *DBA* is equal to the angle *FBC*;

therefore the triangle *ABD* is equal to the triangle *FBC*. [I. 4.

Now the parallelogram *BL* is double of the triangle
*ABD*, because they are on the same base *BD*, and between
the same parallels *BD*, *AL*. [I. 41.

And the square *GB* is double of the triangle *FBC*, because
they are on the same base *FB*, and between the same
parallels *FB*, *GC*. [I. 41.

But the doubles of equals are equal to one another. [*Ax*. 6.

Therefore the parallelogram *BL* is equal to the square *GB*.

In the same manner, by joining *AE*, *BK*, it can be
shewn, that the parallelogram *CL* is equal to the square *CH*.

Therefore the whole square *BDEC* is equal to the two
squares *GB*, *HC'. [*Axiom* 2. *
And the square

*BDEC*is described on

*BC*, and the squares

*GB*,

*HC*on

*BA*,

*AC*

Therefore the square described on the side

*BC*is equal to the squares described on the sides

*BA*,

*AC*.

Wherefore, *in any right-angled triangle* &c. q.e.d.

*THEOREM*.

*If the square described on one of the sides of a tri-angle be equal to the squares described on the other two sides of it, the angle contained by these two sides is a right angle*.

Let the square described on *BC*, one of the sides of
the triangle *ABC*, be equal to the squares described on
the other sides *BA*, *AC*: the angle *BAC* shall be a right angle.

From the point *A* draw *AD* at
right angles to *AC* ; [I. 11.

and make *AD* equal to *BA*; [I. 3.

and join *DC*.

Then because *DA* is equal to
*BA*, the square on *DA* is equal to
the square on *BA*.

To each of these add the square
on *AC*.

Therefore the squares on *DA*, *AC* are equal to the squares on *BA*,*AC* [*Axiom* 2.

But because the angle *DAC* is a right angle, [*Construction*.

the square on DC is equal to the squares on *DA*, *AC*. [I. 47.

And, by hypothesis, the square on *BC* is equal to the squares on *BA*, *AC*.

Therefore the square on *DC* is equal to the square on *BC*. [*Ax* 1.

Therefore also the side *DC* is equal to the side *BC*.

And because the side *DA* is equal to the side *AB*; [Constr.

and the side *AC* is common to the two triangles *DAC*, *BAC*;

the two sides *DA*, *AC* are equal to the two sides *BA*, *AC*, each to each;

and the base *DC* has been shewn to be equal to the base *BC*;

therefore the angle *DAC* is equal to the angle *BAC*. [I. 8.

But *DAC* is a right angle; [*Construction*.

therefore also *BAC* is a right angle. [*Axiom* 1.

Wherefore, *if the square* &c. q.e.d.